To determine the total work done by the boy, let's break down the problem using the given data and steps.

1. Understand the Total Mass:
   The boy and the bicycle have a combined mass: $m = 74 \, \text{kg} + 6 \, \text{kg} = 80 \, \text{kg}$

2. Relationship Between Work and Acceleration:
   The work done can be calculated from the kinetic energy, $W = \Delta KE$. Since the boy starts from rest, we can calculate the work done by integrating the force over the distance $x$. Given that $F = ma$, we can integrate $ma$ over the displacement $x$ using the acceleration provided in the graph.

3. Calculate the Area Under the Acceleration vs. Distance Graph:
   The area under the acceleration graph with respect to distance represents the change in velocity squared (due to $v^2 = 2a \Delta x$), which can then be used to find the work done.

   The graph has a trapezoidal shape. We can divide it into simpler shapes (triangles and a rectangle) and calculate their areas to determine the total "effective acceleration" over the distance.

   • Triangle (0 to 10 m): Area = $\frac{1}{2} \times 10 \, \text{m} \times 2.5 \, \text{m/s}^2 = 12.5 \, \text{m}^2/\text{s}^2$
   • Rectangle (10 to 20 m): Area = $10 \, \text{m} \times 2.5 \, \text{m/s}^2 = 25 \, \text{m}^2/\text{s}^2$
   • Triangle (20 to 30 m): Area = $\frac{1}{2} \times 10 \, \text{m} \times 2.5 \, \text{m/s}^2 = 12.5 \, \text{m}^2/\text{s}^2$

   Total area = $12.5 + 25 + 12.5 = 50 \, \text{m}^2/\text{s}^2$.

4. Calculate the Work Done:
   This area represents $v^2 = 2a \Delta x$, so: $v^2 = 50 \, \text{m}^2/\text{s}^2$

   Thus, the kinetic energy (and hence the work done) is: $W = \frac{1}{2}mv^2 = \frac{1}{2} \times 80 \, \text{kg} \times 50 \, \text{m}^2/\text{s}^2 = 2000 \, \text{J} = 4.0 \, \text{kJ}$

5. Answer:
   The correct answer is C. 4.0 kJ.

Would you like further explanation on any step?

Here are some related questions for further understanding:

1. How would the answer change if the mass of the boy or bicycle were different?
2. What would happen to the work done if the graph showed constant acceleration instead of varying?
3. Can you explain why the area under the acceleration-distance graph relates to the change in velocity squared?
4. How would the work done differ if the boy were riding up an incline?
5. What additional forces (like friction or air resistance) could alter the total work done?

Tip: When interpreting acceleration vs. distance graphs, remember that the area under the curve can relate to velocity, as opposed to acceleration vs. time graphs, where area relates to velocity directly.